Chapter U4.25, Problem 3E

(a)

Interpretation Introduction

Interpretation:

The grams of NaCl required to make PbCl₂ should pre predicted.

Concept Introduction:

The amount of the substance that contains the same number of particles or atoms or molecules is termed as mole.

Answer to Problem 3E

12.6 g of NaCl is required.

Explanation of Solution

The given reaction is represented as follows:

  $2\text{NaCl}\left(aq\right)+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)\to 2{\text{NaNO}}_{\text{3}}\left(aq\right)+{\text{PbCl}}_{\text{2}}\left(s\right)$

The amount of NaCl required is represented as follows:

  $30\text{g}{\text{PbCl}}_{\text{2}}\times \frac{\text{1}\text{mol}{\text{PbCl}}_{\text{2}}}{\text{278}\text{.1}\text{g}{\text{PbCl}}_{\text{2}}}\times \frac{\text{2}\text{mol}\text{NaCl}}{\text{1}\text{mol}{\text{PbCl}}_{\text{2}}}\times \frac{\text{58}\text{.44}\text{g}\text{NaCl}}{\text{1}\text{mol}\text{NaCl}}=12.6\text{g}\text{NaCl}$

Thus, 12.6 g of NaCl is required.

(b)

Interpretation Introduction

Interpretation:

The grams of NaCl required to make PbCl₂ should pre predicted.

Concept Introduction:

The amount of the substance that contains the same number of particles or atoms or molecules is termed as mole.

Answer to Problem 3E

19.8 g of NaCl is required to react with 56 g of lead nitrate. And 47.02 g of PbCl₂ is formed.

Explanation of Solution

The given reaction is represented as follows:

  $2\text{NaCl}\left(aq\right)+\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_2\left(aq\right)\to 2{\text{NaNO}}_{\text{3}}\left(aq\right)+{\text{PbCl}}_{\text{2}}\left(s\right)$

The amount of NaCl required is represented as follows:

  $56\text{g}\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\times \frac{\text{1}\text{mol}\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{331}\text{.2}\text{g}\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\times \frac{\text{2}\text{mol}\text{NaCl}}{\text{1}\text{mol}\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\times \frac{\text{58}\text{.44}\text{g}\text{NaCl}}{\text{1}\text{mol}\text{NaCl}}=19.8\text{g}\text{NaCl}$

Thus, 19.8 g of NaCl is required.

Now, the grams of PbCl₂formed is represented as follows:

  $56\text{g}\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\times \frac{\text{1}\text{mol}\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{331}\text{.2}\text{g}\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\times \frac{\text{1}\text{mol}{\text{PbCl}}_2}{\text{1}\text{mol}\text{Pb}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}\times \frac{278.1\text{g}{\text{PbCl}}_2}{\text{1}\text{mol}{\text{PbCl}}_2}=\text{47}\text{.02g}{\text{PbCl}}_2$

Thus, 47.02 g of PbCl₂is formed.